To be more clear, I post this 'unnecessary optimization' of Vegaseat's code in new thread instead of end of old one. Just to show that this is even simpler than checking if given day is first weekday of the month.
List of first work days of month for any year
"""Task: calculate this list (from vegaseat
http://www.daniweb.com/software-development/legacy-and-other-languages/threads/362098/1554079#post1554079)
All first workdays of each month in 2011:
1) Mo 01/03/2011
2) Tu 02/01/2011
3) Tu 03/01/2011
4) Fr 04/01/2011
5) Mo 05/02/2011
6) We 06/01/2011
7) Fr 07/01/2011
8) Mo 08/01/2011
9) Th 09/01/2011
10) Mo 10/03/2011
11) Tu 11/01/2011
12) Th 12/01/2011
"""
from datetime import date, timedelta
def first_workday(month, year):
first = date(year, month, 1)
if first.weekday() < 5:
return first
return first + timedelta(days=7-first.weekday())
year = 2011
for month in range(1,13):
print(' %2i) %s' % (month, first_workday(month,year).strftime('%a %m/%d/%Y')))
# Produces three letter weekday name instead of two letters
TrustyTony
888
pyMod
Team Colleague
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Not only way to do it of course (in spite of Python philosophy):
"""Task: calculate the list from vegaseat
http://www.daniweb.com/software-development/legacy-and-other-languages/threads/362098/1554079#post1554079)
"""
from datetime import date, timedelta
def first_workday(month, year):
first = date(year, month, 1)
return first if first.weekday() < 5 else first.replace(day=8 - first.weekday())
def table(year = 2011):
print('\n'.join(' %2i) %s' %
(month, first_workday(month,year).strftime('%a %m/%d/%Y'))
for month in range(1,13)))
# Produces three letter weekday name instead of two letters
table()
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